Understanding Grounding and Induced Charges on Concentric Spherical Conductors

[meteorologist1]

Hi, I would like to ask a question about grounding conductors. Suppose we have concentric spherical thin conducting shells (consisting of an inner conductor and an outer conductor). Suppose a charge of +Q is placed on the inner conductor. If we ground the outer conductor, my understanding is that there will be a charge of -Q induced on it. But, what happens if the inner conductor is grounded while a charge of +Q is placed on the outer conductor? Could someone explain to me this phenomenon between grounding and induced charges? Thanks.

 

[MathStudent]

The ground is just a wire connecting the conducor to the ground (as in a pole stuck in the Earth or some other neutral object). This acts as an infinite source, or conversely, an infinite sinkhole for electrons. You can't ground the inner conductor because it is encapsulated by the outer one... thus you would have to drill a hole in the outer and put some kind of insulator between the ground and the outer conductor so as to not permit the transfer of electrons from outer to inner. By the way... the ground just gives the charges an outlet in which to move...

As an example of induced charge:
Consider a neutral metal block with a wire connecting one end to a pole stuck in ground. Now bring a distribution of charge, like a negatively charged metal wand, close to the end of the block that isn't connected to the wire. The electrons in the block will be repelled by the negative charge of the wand and will be forced to travel down the wire to the ground.
Now this is the important part:
WHILE the wand is still held CLOSE (but not touching, we don't want charges to transfer from the wand to the block ) disconnect the wire to the ground, thus severing any way for charges to move to or from the block. Then when you move the wand away the block will be left with a net positive charge. This is becase some of the electrons orginally present in the neutral block were reppelled by the wand and traveled to the ground when the wand was brought close. however since we disconnected the ground,,, those electrons were unable to return to the block once the source of repulsion was removed. Now the block has fewer electrons than it did before, leaving a net positive charge. Therefore, the wand induced a positive charge in the block...
Does this make any sense.
Had we left the ground connected to the block, the electrons would have returned to the block once the wand was taken away giving you a neutral block.

PS... had to edit out a few "thus's" I tend to use that word a lot

 

[meteorologist1]

Ok I see. Thanks for your explanation. But this is a problem I'm doing in my class -- just suppose we could ground the inner conductor. And suppose the outer conductor has charge +Q. What would the electric field and potentials be inside the inner conductor, between the two conductors, and outside the outer conductor? Based on your explanation, it seems that the grounding has no effect on the inner conductor. It would have charge zero?

 

[MathStudent]

Could you write the question as its written in your book? I am not sure about what's going on with the ground (is it disconnected at any point)

 

[meteorologist1]

The question is:
Consider the problem of concentric spherical thin conducting shells, inner conductor r=a, outer conductor r=b, where a<b. Assume ideal conductors. Using Gauss, find expressions for E and V everywhere, and sketch each as a function of r, for the case of a charge +Q on the outer conductor and the inner conductor grounded.

Right now I'm thinking that the answers would be E=0 for r<a and a<r<b, and E nonzero for r>b. I'm not sure about the potential V; it should be constant for r<a and a<r<b; and nonconstant for r>b. On the other hand, doesn't ground mean at r=a, V=0? If so, there is a discontinuity, which is not good.

 

[MathStudent]

The ground is what's throwing me off because it seems unnecessary I believe the inner Sphere stays neutral regardless of whether it's there or not.
Essentially the inner sphere doesn't effect the situation at all.
Does anyone see a problem with this?
The outer sphere would act as a point charge located at the center of the sphere with charge +Q. Therefore You would expect E and V to have the same values (for r>b) as that of a point charge with charge +Q.

However they want you to prove this using Gauss's law which is easily done. let your Gaussian surface be a sphere with radius > b hence E is perpendicular to the surface at all points and is uniform so it can be taken out of the integral, thus you just need the formula for the surface area of a sphere.

remember Gauss's Law is:
$$\oint\vec{E}\cdot\vec{dA} = \frac{Q_{encl}}{\epsilon_0}$$

 

[meteorologist1]

Maybe the grounding of the inner conductor is used to force you to set the potential at r=a to be 0. Therefore, V=0 in r<a and a<r<b, and V will be negative for r>b in order to keep V(r) continuous.

 

[MathStudent]

what makes you think the ground makes the potential at r=a to be zero. In order for the potential to change there needs to be an electric field... since the inner sphere is not charged, then there is no electric field inside the outer sphere... therefore the potential is constant inside the sphere with radius = b

 

[meteorologist1]

Ok I see. What you said makes sense. I just thought that the "definition" for ground is that the potential there is zero, but nevermind.

 

[MathStudent]

The potential is a property of space due to charges, therefore it is possible to have a point in space where there is zero electric field but there is some potential > or < 0

 

[maverick280857]

what makes you think the ground makes the potential at r=a to be zero. In order for the potential to change there needs to be an electric field... since the inner sphere is not charged, then there is no electric field inside the outer sphere... therefore the potential is constant inside the sphere with radius = b

Are you saying that grounding a conductor may not make it's potential equal to zero??

 

[da_willem]

the potential is not determined up to a constant. Only differences in the potential (electric fields) matter. For example you could say the potential of the Earth (ground) is 10000000000V then the potential of your power takeoff is 10000000230 V, and all your electronic devices still work. Usually one takes the potential of the ground as zero though.

($$\vec{E}=\nabla V=\nabla (V+cst)$$. This allowed adding of a constant to the potential is called 'gauge freedom'))

 

[maverick280857]

the potential is not determined up to a constant. Only differences in the potential (electric fields) matter. For example you could say the potential of the Earth (ground) is 10000000000V then the potential of your power takeoff is 10000000230 V, and all your electronic devices still work. Usually one takes the potential of the ground as zero though.

($$\vec{E}=\nabla V=\nabla (V+cst)$$. This allowed adding of a constant to the potential is called 'gauge freedom'))

Yes to be sure that's what I meant (usually $$R_{earth}$$ is large enough to enable us--for practical purposes--to take $$V_{earth} = 0$$). But why is this called "gauge freedom"? Thats interesting...

 

[da_willem]

This was more of a general comment on the scalar potential. To clarify doubtful statements like "The potential is a property of space"...

About your remark:

Are you saying that grounding a conductor may not make it's potential equal to zero??

grounding a conductor gives the conductor (practically) the same potential as your ground. But calling this potential zero is an arbitrary, but ofcourse practical choice.

 

[MathStudent]

This was more of a general comment on the scalar potential. To clarify doubtful statements like "The potential is a property of space"...

Let me clarify on my previous remarks:
I was trying to point out that a change in potential is dependent upon the electric field. I wasn't trying to imply that the Earth ( or ground) has a definitive potential value. Of course this value is arbitrary, like you said only the difference in potential is what matters. I failed to state that you can define the potential of a point to be any value you want and then must consider other values relative to this definition.

But are you implying that potential is not a property of space? For a given charge distribution there is a numerical value for potential at every point in space wrt some defined defined value at a given point.

 

[maverick280857]

No I tend to agree with what you say now (I was unable to understand the implication in your very first post though). And yes, its a choice to take V(earth) as zero...I could well take it as a finite quantity as in the difference of potentials, this would cancel out just as a reference potential does (and we need not take it as infinity always).

 

[da_willem]

But are you implying that potential is not a property of space? For a given charge distribution there is a numerical value for potential at every point in space wrt some defined defined value at a given point.

I'm not sure if you should call the potential (maybe not even the fields) a property of space. But this is becoming more philiosophy than physics. So I'll let you return to your topic.

 

[maverick280857]

But this is becoming more philiosophy than physics. So I'll let you return to your topic.

Why do you think so mate?

 

[da_willem]

Why do you think so mate?

Before Einsteins special relativity, they though electromagnetic waves were some sort of distortion of space, a ripple in the continuous background of space, the ether. The transormation formula for time and space made the assumption of an ether unnecessary. The general consensus among physicists is that electromagnetic fields are entities in itself and electromagnetic waves unlike mechanical waves not distortions in some sort of medium.

 

[maverick280857]

Now, I know very little about physics so I cannot comment on that but I would like to believe that a simple visual picture about the original question would help and at the same time would some ideas about electromagnetism.

 

[Gokul43201]

Maybe the grounding of the inner conductor is used to force you to set the potential at r=a to be 0.

This is correct.

Therefore, V=0 in r<a

Yes. But this does not automatically follow from the above, unless you also use Gauss' Law.

and a<r<b,

No, (if you mean V=0 in between the shells) why should this be true ? Surely this would at least violate continuity at r=b, where you suddenly find a non-zero potential.

Use Gauss' Law to find $$E(a<r<b) = \frac{-Q}{4 \pi \epsilon _0 r^2} = - \nabla V(r)$$

and V will be negative for r>b in order to keep V(r) continuous.

No, how does V < 0 ensure continuity ?

Clearly E = 0, outside, from Gauss. And V > 0 at r=b, as you will find if you complete the previous part. From this, you can find V(r>b).

 

[MathStudent]

No, (if you mean V=0 in between the shells) why should this be true ? Surely this would at least violate continuity at r=b, where you suddenly find a non-zero potential.

By saying that V does not = 0 in between the shells, you are also saying that there is some nonzero E with a radial component in between the shells. How do you explain this E?
What I got from the problem is that there is no charge on the inner sphere, and there is only charge on the outer sphere, so there is no Q enclosed by any gaussian surface with r<b ( I figure if I am wrong anywhere it is probably here... but I just can't seem to argue how any charge could exist on or within the inner sphere).
Taking this into account, and due to the symmetry of the problem, you can say that there is no E anywere within the system of conductors with r<b ( by Gauss's law ). Am I missing something?... If I am wrong, then please correct me.
Thanks
-MS

 

[JFo]

By saying that V does not = 0 in between the shells, you are also saying that there is some nonzero E with a radial component in between the shells. How do you explain this E?
What I got from the problem is that there is no charge on the inner sphere, and there is only charge on the outer sphere, so there is no Q enclosed by any gaussian surface with r<b ( I figure if I am wrong anywhere it is probably here... but I just can't seem to argue how any charge could exist on or within the inner sphere).
Taking this into account, and due to the symmetry of the problem, you can say that there is no E anywere within the system of conductors with r<b ( by Gauss's law ). Am I missing something?... If I am wrong, then please correct me.
Thanks
-MS

I think your right, there wouldn't be any charge on the inner sphere.

 

[Gokul43201]

What I got from the problem is that there is no charge on the inner sphere

Here's where I believe you're mistaken.

The act of grounding a conductor does not make it charge free. If the negative charges are free to move to the earth, why don't they ? Because they have a greater motivation for wanting to stay on the conductor. This motivation is created by the conductor with a positive charge on it.

Consider the 2 parallel plates (easier to draw), one of which (the one on the right) has an excess charge, +Q.

If the second plate is not grounded, it will experience charge separation, whereby a negative charge is induced on the side nearer the other plate, and a positive charge on the far side. This makes sense : opposite charges attract, etc. But the net charge on the second plate remains zero because no charge is added to or removed from this plate.

Now if you connect plate II to ground, you are providing a path for the induced positive charges to move through, thus making themselves as far away from the first (positively charged) plate as possible. {In reality, since positive charges don't move in conventional conductors, what you have is negative charge (electrons) rushing up from ground, to neutralize the positive charges.} What's left on the second plate is the induced negative charge -Q.

     II            I                      II           I 
  ________      ________               ________      ________
  |+    -|      |      |               |     -|      |      |
  |+    -|      |      |               |     -|      |      |
  |+ 0  -|      | +Q   |               | -Q  -|      | +Q   |
  |+    -|      |      |               |     -|      |      |
  |+    -|      |      |               |     -|      |      |
  --------      --------               --------      --------
                                         |
                                       __|__
                                        ---
                                      Ground

 

[MathStudent]

Gokul43201, Thank you for your detailed response ( I hope you didn't spend too much time on the diagram, it is appreciated)

This makes sense for the case of two parallel plates, but I don't think this same reasoning applies to the case of spherical shells due to its symmetry.

For example:
Consider just a spherical metal conducting shell with a Charge of +Q evenly distributed on its surface. This obviously sets up an electric field for all points outside the shell, but for all points inside, the electric field is zero (by Symmetry and Gauss's Law). I would immagine then that if we were to place another (neutral) conducting shell inside the charged shell, then there would be no tendency towards a charge separation since there is no electric field compelling the free charges to move.

In the case of the parallel plates, the charge separation was caused by the electric field set up by the + plate. But in this case, there is no electric field anywhere inside the outer shell.

 

[Gokul43201]

This makes sense for the case of two parallel plates, but I don't think this same reasoning applies to the case of spherical shells due to its symmetry.

You are perfectly right !

The potential will be zero anywhere inside the r=b. I must be losing it !